Let \(A\subseteq\mathbb N\). Then, we define the set of sums of finitely many elements from \(A\).
$$\operatorname{FS}(A) = \left\{\sum_{i=0}^na_i\quad:\quad n\in\mathbb N,\ \ a_i\in A,\ \ \text{ the \(a_i\) are all different}\right\}$$
Note that \(\mathrm{FS}(A)\) is not closed under finite sums. For instance, \[\operatorname{FS}(\{1\}\cup3\mathbb N)=3\mathbb N\cup(3\mathbb N+1)\]
For allcolorings of the integers in a finite number of colors \(\mathbb N=C_0\cup\cdots\cup C_{n-1}\), there exists a color \(i\) and an infinite set \(H\subseteq \mathbb N\) and such that \[\operatorname{FS}(H)\subseteq C_i\]
The coloring of even and odd integers,
The \(\mathrm{mod}\ 3\) coloring,
The coloring given by the parity of the length of the binary expansion
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
...
A more general formulation
An IP set is a set \(A\subseteq\mathbb N\) such that there exists an infinite \(S\) such that \(\operatorname{FS}(S)\subseteq A\)
\(\mathbb N\) is an IP set, but no finite set in an IP set,
If \(A\) is an IP set and \(B\supseteq A\), then \(B\) is an IP set.
A class \(\mathcal C\subseteq\mathcal P(\mathbb N)\) is partition regular, if for every \(A\in\mathcal C\), if \(A\) is partitionned in \(A=A_0\cup\dots\cup A_{n-1}\), then for some \(i\lt n\), \(A_i\in\mathcal C\).
The IP sets form a partition regular class. For anyIP set \(A=A_0\cup\cdots\cup A_{n-1}\), for some \(i\lt n\), \(A_i\) is an IP set.
This reveals the "Ramseyan" statement: for every coloring of a structure, there is monochromatic substructure
This allows to iterate Hindman's Theorem!
State of the art of the Reverse Mathematics of Hindman's theorem.
For every finite coloring \(c:\mathbb N\to2\), there exists a monochromatic IP set \(H\) computable in the \((\omega+1)\)-th Turing jump of \(c\).
Hindman Theorem is provable in \(\mathrm{ACA}_0^+\).
There exists a computable finite coloring such that all homogeneous sets \(H\) compute \(\emptyset'\), and one with no \(\Sigma^0_2\) solution.
Hindman's theorem proves \(\mathrm{ACA}_0\).
Given a computable coloring \(c\):
\(\emptyset^{(\omega+1)}\) computes a monochromatic IP set.
\(\emptyset^{'}\) might not compute a monochromatic IP set.
What is the exact \(\alpha\) such that:
\(\emptyset^{(\alpha)}\) computes a monochromatic IP set,
\(\emptyset^{(\beta)}\) might not compute a monochromatic IP set for \(\beta<\alpha\)?
In terms of reverse mathematics:
Is Hindman Theorem provable in \(\mathrm{ACA}_0\)?
(~corresponds to \(\alpha\lt\omega\))
Does Hindman Theorem proves in \(\mathrm{ACA}^+_0\)?
(~corresponds to \(\alpha\geq\omega\))
Is it strictly in between?
(due to lack of induction)
Effective proof of Hindman Theorem
Define a largeness notion such that we can prove: $$\text{The large sets form a partition regular class}\tag{1}$$ $$A\text{ is large }\Longrightarrow \exists n\in A, A\cap A+n\text{ is large. }\tag{2}$$
What are the possible candidate for this notion of largeness? In particular, a large set must be an IP set!
(2) does not hold for large being infinite
(1) is equivalent to Hindman's Theorem when large is being an IP set
Example of largeness notions that work
Define a largeness notion such that we can prove: $$\text{The large sets form a partition regular class}\tag{1}$$ $$A\text{ is large }\Longrightarrow \exists n\in A, A\cap A+n\text{ is large. }\tag{2}$$
An idempotent ultrafilter is an ultrafilter \(\mathcal U\) such that \(\mathcal U = \mathcal U + \mathcal U\), where \[A\in\mathcal U + \mathcal U \Longleftrightarrow \{n:A-n\in\mathcal U\}\in\mathcal U\]
For any idempotent ultrafilter \(\mathcal U\),
If \(A=A_0\cup\dots\cup A_{n-1}\in\mathcal U\), then \(\exists i\lt n\) s.t. \(A_i\in\mathcal U\),
If \(A\in\mathcal U\), then \(\exists n\in A\) s.t. \(A\cap A+n\in\mathcal U\).
If \(I\) is an IP set, a set \(A\) is said to be large in \(I\) if there is no IP set \(I'\subseteq I\), such that \(I'\cap A=\emptyset\). It is B-large if it is large in some IP set \(I\).
If \(A\) is large in \(I\), there is no way to avoid color \(A\) by shrinking \(I\)!
If \(A=\bigcup_{i\lt n}A_i\) is B-large, then \(\exists i\lt n\) s.t. \(A_i\) is B-large,
If \(A\) is B-large, then \(\exists n\in A\) s.t. \(A\cap A+n\) is B-large.
As a consequence, a set is B-large if and only if it is an IP set.
Effectivizing the proof
Recall
$$A\text{ is large }\Longrightarrow \exists n\in A, A\cap (A+n)\text{ is large. }\tag{2}$$
The problem is that the \(n\) found by (2) is chosen in a non uniform way, out of infinitely many, and iterated \(\omega\) many times. Consider:
$$A\text{ is large }\Longrightarrow \exists {\color{maroon}n_0,\cdots, n_N}\in A, \bigcup_{i\lt N}A\cap (A+{\color{maroon}n_i})\text{ is large. }\tag{2'}$$
We have that \((1)+(2')\Rightarrow(2)\):
$$A\text{ is large }\xRightarrow{(2')}\bigcup_{n\in F}A\cap (A+n)\text{ is large }\xRightarrow{(1)}\exists n\in F, A\cap (A+n)\text{ is large.}$$
This time, \(n\) is found by (1) in a non uniform way, but out of finitely many. We postpone this choice to the end, and split the construction.
Avoiding the use of (1) in the proof, supposing (2')
$$A\text{ is large }\Longrightarrow \exists {\color{maroon}n_0,\cdots, n_N}\in A, \bigcup_{i\lt N}A\cap (A+{\color{maroon}n_i})\text{ is large. }$$
A full-match for a coloring \(c\) is a couple \(({\color{maroon}F},S)\) where \(F\) is a finite set, \(S\) is an IP set, and for every \(x\in S\), there exists \({\color{maroon}a}\in {\color{maroon}F}\) such that \[c({\color{maroon}a})=c(x)=c({\color{maroon}a}+x)\]
The complexity of a full-match is essential to understand the complexity of a solution to Hindman's theorem!
Complexity of full/right/left-matches
A full-match for a coloring \(c\) is a couple \(({\color{maroon}F},S)\) where \(F\) is a finite set, \(S\) is an IP set, and for every \(x\in S\), there exists \({\color{maroon}a}\in {\color{maroon}F}\) such that \[c({\color{maroon}a})=c(x)=c({\color{maroon}a}+x)\]
Does there always exists a computable full-match given a computable coloring?
An affirmative answer would prove Hindman's theorem equivalent to \(\mathrm{ACA}_0\)
A negative answer to the question was published, but there was an error in the proof.
The proof of the existence of a full-match is similar to Hindman's Theorem, given the existence of right-matches. The proof yields an arithmetical solution.
A right-match for a coloring \(c\) is a couple \((F,S)\) where \(F\) is a finite set, and for every \(x\in\operatorname{FS}(S)\), there exists \(a\in F\) such that \[c(x)=c(a+x)\]
Every computable coloring admits a computable right-match, uniformly in the double jump of the coloring.
A left-match for a coloring \(c\) is a couple \((F,S)\) where \(F\) is a finite set, and for every \(x\in\operatorname{FS}(S)\), there exists \(a\in F\) such that \[c(a)=c(a+x)\]
Every computable coloring in two colors admits a computable left-match.
However, the following hold:
The existence of computable full-match for 2 colors implies the existence of computable full-matches for \(n\) colors, but
The existence of computable left-match for 2 colors does not imply the existence of computable left-matches for \(n\) colors.
Does there always exist a computable full-match given a computable 2-coloring?
Does there always exist a computable left-match, given a computable 3-coloring?
Consider the following property:
$$\exists F\subseteq C_i, \forall x\in\operatorname{FS}(S),\ \exists a\in F,\ c(a+x) = j\tag{\(\mathrm{MM}_{i,j}\)}$$
Every coloring in two colors satisfying one of the following properties:
\(\lnot(\mathrm{MM}_{0,1})\) or \(\lnot(\mathrm{MM}_{1,0})\)
\((\mathrm{MM}_{0,0})\) and \((\mathrm{MM}_{1,1})\)
\(\lnot(\mathrm{MM}_{0,0})\) and \(\lnot(\mathrm{MM}_{1,1})\)